Calculus IV Quiz 1 Fall 1998 9/9/98

The idea is that you hold x constant and take the derivative with y as the variable, but the definition is f_y =.
 
 
2. Write an equation for the plane tangent to the surface z = y³ - 2xy at the point (1, -2).

First we find the two partial derivatives:
f_x = -2y
f_y = 3y² - 2x

Then we evaluate them at (1, -2):
f_x(1, -2) = -2(-2) = 4
f_y(1, -2) = 3(-2)² - 2(1) = 10

We also need a z coordinate for our point of tangency, so we plug (1, -2) into our original function:
z₀ = (-2)³ - 2(1)(-2) = -4

And then we pop all these things into the appropriate slots in the point-slope equation for a plane:
z - z₀ = f_x(x₀, y₀) (x - x_o) + f_y(x₀, y₀) (y - y₀)
z + 4 = 4(x-1) + 10(y+2)

We can either leave it like that (since the question didn't specify a particular form for the answer) or carefully simplify to something like:
z = 4x + 10y + 12
 
 
3. Let f(x,y) = . Find the gradient of f at the point (1, -1).

As usual, we first need to find the partial derivatives with respect to x and y:

Then we evaluate each of these at the point (1, -1):
f_x(1, -1) = -2(1)(-1)e^-2 = 2e^-2
f_y(1, -1) = e² + -2(-1)² e^-2 = -e^-2

And finally we put these together as the x and y components of the gradient vector:
grad f = <2e^-2,-e^-2>